Problem: You are given an integer N.

Construct any one N-by-N matrix a that satisfies the conditions below. It can be proved that a solution always exists under the constraints of this problem.

* 1 \leq a_{i,j} \leq 10^{15}
* a_{i,j} are pairwise distinct integers.
* There exists a positive integer m such that the following holds: Let x and y be two elements of the matrix that are vertically or horizontally adjacent. Then, {\rm max}(x,y) {\rm mod} {\rm min}(x,y) is always m.

Constraints

* 2 \leq N \leq 500

Input

Input is given from Standard Input in the following format:


N


Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Output

Print your solution in the following format:


a_{1,1} ... a_{1,N}
:
a_{N,1} ... a_{N,N}

Example

Input

2


Output

4 7
23 10

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
LL read(){
	LL x=0;
	char ch=getchar();
	while (!isdigit(ch))
		ch=getchar();
	while (isdigit(ch))
		x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
	return x;
}
const int N=505;
int n;
vector <int> vp;
LL g[N][N];
LL gcd(LL x,LL y){
	return y?gcd(y,x%y):x;
}
LL lcm(LL x,LL y){
	if (!x||!y)
		return x+y;
	return x/gcd(x,y)*y;
}
int check(int n){
	for (int i=2;i*i<=n;i++)
		if (n%i==0)
			return 0;
	return 1;
}
map <int,int> mp1,mp2;
int g1(int x){
	if (!mp1[x])
		mp1[x]=vp.back(),vp.pop_back();
	return mp1[x];
}
int g2(int x){
	if (!mp2[x])
		mp2[x]=vp.back(),vp.pop_back();
	return mp2[x];
}
int main(){
	n=read();
	if (n==2){
		puts("4 7");
		puts("23 10");
		return 0;
	}
	for (int i=2;vp.size()<n*2;i++)
		if (check(i))
			vp.push_back(i);
	reverse(vp.begin(),vp.end());
	mp1.clear(),mp2.clear();
	for (int i=1;i<=n;i++)
		for (int j=1;j<=n;j++)
			if ((i+j)%2==0)
				g[i][j]=g1(i+j);
	for (int i=n;i>=1;i--)
		for (int j=n;j>=1;j--)
			if ((i+j)%2==0)
				g[i][j]*=g2(i-j);
	for (int i=1;i<=n;i++)
		for (int j=1;j<=n;j++)
			if ((i+j)%2==1){
				g[i][j]=lcm(g[i-1][j],lcm(g[i+1][j],lcm(g[i][j-1],g[i][j+1])))+1;
				assert(g[i][j]<=1000000000000000LL);
			}
	for (int i=1;i<=n;i++,puts(""))
		for (int j=1;j<=n;j++)
			printf("%lld ",g[i][j]);
	return 0;
}